# Proving the trigonometric identity $4\sin\left(x + \frac{\pi}{6}\right)\sin\left(x - \frac{\pi}{6}\right) = 3 - 4 \cos^2 x$

I am using the addition and difference formulae for cosine and sine but seem to be getting stuck... somewhere.

Prove that

$$4\sin\left(x + \frac{\pi}{6}\right)\sin\left(x - \frac{\pi}{6}\right) = 3 - 4 \cos^2 x.$$

• It would help for us to see how far you’ve gotten in this. – Lubin Jan 18 '15 at 3:02

## 3 Answers

We can safely use, Prove $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$

$$\sin\left(x+\dfrac\pi6\right)\sin\left(x-\dfrac\pi6\right)=\sin^2x-\sin^2\dfrac\pi6=1-\cos^2x-\left(\dfrac12\right)^2=\cdots$$

Alternatively

observe that $\cos3A=4\cos^3A-3\cos A=\cos A(4\cos^2A-3)$

Use the pattern described in Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,

to arrive at $4\cos3A=-\cos A\sin\left(A+\dfrac\pi6\right)\sin\left(A-\dfrac\pi6\right)$

\begin{align} 4 \sin(x+\frac{\pi}6)\sin(x-\frac{\pi}6) &= 4 [\sin(x)\cos(\frac{\pi}6)+\cos(x)\sin(\frac{\pi}6)][\sin(x)\cos(\frac{\pi}6)-\cos(x)\sin(\frac{\pi}6)] \\ &=4[\sin^2(x)\cos^2(\frac{\pi}6)-\cos^2(x)\sin^2(\frac{\pi}6)] \\ &=4[\sin^2(x)\frac 34-\cos^2(x)\frac 14] \\ &=3\sin^2(x)-\cos^2(x) \\ &=3(1-\cos^2(x))-\cos^2(x) \\ &=3-3\cos^2(x)-\cos^2(x) \\ &=3-4\cos^2(x) \end{align}

Take $$\theta = 2x , \qquad \phi = \frac{\pi}{3}$$ in the sum-to-product identity $$\cos \theta - \cos \phi = -2 \sin \left(\frac{\theta + \phi}{2}\right) \sin \left(\frac{\theta - \phi}{2}\right).$$

Multiplying both sides by $-2$ and then invoking the double-angle identity $\cos 2x = 2 \cos^2 x - 1$ gives \begin{align}4 \sin \left(x + \frac{\pi}{6}\right) \sin \left(x - \frac{\pi}{6}\right) &= -2\left(\cos 2x - \cos \frac{\pi}{3}\right) \\ &= -2 \left[ 2 \cos^2 x - 1 - \frac{1}{2}\right] \\ &= 3 - 4 \cos^2 x .\end{align}